#includevoid put(int ar[][5],int n);int main(void){int zhou[3][5]={{1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7}};put(zhou,5);return 0;}void put(int ar[][5],int n){int r,c;for(r=0;r
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![#includevoid put(int ar[][5],int n);int main(void){int zhou[3][5]={{1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7}};put(zhou,5);return 0;}void put(int ar[][5],int n){int r,c;for(r=0;r](/uploads/image/z/8768719-55-9.jpg?t=%23includevoid+put%28int+ar%5B%5D%5B5%5D%2Cint+n%29%3Bint+main%28void%29%7Bint+zhou%5B3%5D%5B5%5D%3D%7B%7B1%2C2%2C3%2C4%2C5%7D%2C%7B2%2C3%2C4%2C5%2C6%7D%2C%7B3%2C4%2C5%2C6%2C7%7D%7D%3Bput%28zhou%2C5%29%3Breturn+0%3B%7Dvoid+put%28int+ar%5B%5D%5B5%5D%2Cint+n%29%7Bint+r%2Cc%3Bfor%28r%3D0%3Br)
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#includevoid put(int ar[][5],int n);int main(void){int zhou[3][5]={{1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7}};put(zhou,5);return 0;}void put(int ar[][5],int n){int r,c;for(r=0;r
#include
void put(int ar[][5],int n);
int main(void)
{
int zhou[3][5]={
{1,2,3,4,5},
{2,3,4,5,6},
{3,4,5,6,7}
};
put(zhou,5);
return 0;
}
void put(int ar[][5],int n)
{
int r,c;
for(r=0;r
#includevoid put(int ar[][5],int n);int main(void){int zhou[3][5]={{1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7}};put(zhou,5);return 0;}void put(int ar[][5],int n){int r,c;for(r=0;r
#include
void put(int ar[][5],int n);
int main(void)
{
int zhou[3][5]={
{1,2,3,4,5},
{2,3,4,5,6},
{3,4,5,6,7}
};
put(zhou,3);
return 0;
}
void put(int ar[][5],int n)
{
int r,c;
for(r=0;r
#includevoid put(int ar[][5],int n);int main(void){int zhou[3][5]={{1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7}};put(zhou,5);return 0;}void put(int ar[][5],int n){int r,c;for(r=0;r
c++求2+22+222+2222+22222.#includevoid main(){int sum=0;int a=2;int n;coutn;for(int i=1;i
以下程序的运行结果是输出如下图形** * ** * * * ** * * * * * ** * * * * * * * *#includevoid a(int i){int j,k;for(j=0;j
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简单C++,给每行注释吧.#includevoid main(){ int i=1,j; for(;i
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C++程序解释请详细说明含义好吗?#include#includevoid main(){int* p;p=new int [8];for(int i=0;i
C++求两个集合的交集#includevoid main(){ int a[5],b[5],c[5]; int i,j,k; int *p1,*p2,*p3; cout
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C语言正数相加变负数#includevoid main(){ int i,sum=0;for(i=200;i
#includevoid main(){int *p1,*p2,a[5]={1,3,5,7,9};for(p1=a;p1
#includevoid fun(int x){ if(x/2>0) fun(x/2);printf(%d,x);}main(){fun(3); printf(
);}
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求下列程序输出结果#includevoid main(){int a=3;printf(%d
,(a+=a-=a*a));}